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Amplitude of a block oscillating on a spring

  • Thread starter drunknfox
  • Start date
  • #1
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Homework Statement


a .25kg block oscillates on the end of a spring with a spring constant of 200N/m. If the oscillation is started by elongating spring to 0.15m and giving the block a speed of 3.0m/s. The amplitude is?


Homework Equations


omega= sqrt(k/m), Vm = (omega)Xm


The Attempt at a Solution



omega= sqrt (200/.25) = 28.28 (3m/s) = 28.28Xm (3/28.28) = .106

The correct answer is .18m..what am i missing?
 

Answers and Replies

  • #2
ehild
Homework Helper
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The spring is stretched by 0.15 m and the block has 3 m/s speed at the same time. 0.15 is not the maximum displacement and 3 m/s is not the maximum speed. Calculate the energy, and get the maximum speed from it.

ehild
 

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